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$$g(x)=\begin{cases} x+2x^2\sin\left(\frac{1}{x}\right)&\text{ if }x\neq0\\\ 0&\text{ if }x=0 \end{cases}$$ Show that there is a sequence $\{x_n\}$ with $\{x_n\} \to 0$ as $n$ approaches infinity, such that $g'(x_n)=0\ \forall n$ but $g'(0) \ne 0$.

I calculated $$g'(x)=4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right)+1 ,\text{ if }x\neq0$$

I tried to construct a sequence where $\{x_n\}=\dfrac 1 {2\pi n}$ so that $g'(x_n)=0$, but I'm not sure if this is the right way to prove the question. If this is not right, how can I go about showing it?

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  • $\begingroup$ What is $f$? The same as $g$? $\endgroup$ – miracle173 Nov 20 '15 at 4:19
  • $\begingroup$ @miracle173 Yes, sorry I mistyped it. $\endgroup$ – joyce Nov 20 '15 at 4:22
  • $\begingroup$ And what is $g'$ if $x \ne 0$? Did you already calculate it? $\endgroup$ – miracle173 Nov 20 '15 at 4:23
  • $\begingroup$ I think $g'$ is $4xsin(1/x)-2cos(1/x)+1$ $\endgroup$ – joyce Nov 20 '15 at 4:26
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    $\begingroup$ $g'(x)=\begin{cases} 1+4x\sin\frac{1}{x}-2\cos\frac{1}{x}&\text{ if }x\neq0\\\ 1&\text{ if }x=0 \end{cases}$ $g'(\frac{1}{2n\pi})<0$, $g'(\frac{1}{2n\pi+\pi/2})>0$,By intermediate value theorem $\exists \frac{1}{2n\pi+\pi/2}<x_n<\frac{1}{2n\pi}$, $g'(x_n)=0$. I don't know how to construct the concrete $x_n$ $\endgroup$ – Laura Nov 20 '15 at 4:33
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The first summand of your $g'$, $$4x\sin\left(\frac{1}{x}\right)$$ converges to $0$ if $x$ converges to $0$, because $|4\sin\left(\frac{1}{x}\right)|\le 4$.

$$\cos(\phi)=\frac{1}{2}, \text{iff} \;\phi=\pm\frac{\pi}{3}\pm 2k\pi$$

so choose $$\frac{1}{x}=\pm\frac{\pi}{3}\pm 2k\pi$$ e.g. $$x_n=\frac{1}{\frac{\pi}{3}+ 2n\pi}$$

But in a similar way you can construct sequences $x_n$ that converge to a value different to $0$.

$$g'(0)=\lim_{x \to 0}\left(\frac{g(x)-g(0)}{x-0}\right)=\lim_{x \to 0}\left(1-x\sin\left(\frac{1}{x}\right)\right)=1$$ So the constructed sequence shows that the derivate $g'$ of $g$ is not continous at $x=0$.

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  • $\begingroup$ But if we plug $x_n=\frac{1}{\frac{\pi}{3}+ 2n\pi}$ in to the equation of $$g'(x)=4x\sin\left(\frac{1}{x}\right)-2\cos\left(\frac{1}{x}\right)+1 ,\text{ if }x\neq0$$, we do not get $g'(x_n)=0$, that is we do not have $sin \frac{1}{\frac{\pi}{3}+ 2n\pi}=0$ $\endgroup$ – joyce Nov 20 '15 at 6:28
  • $\begingroup$ you get $g'(x_n)= \frac{4}{\frac{\pi}{3}+ 2n\pi}\sin (\frac{\pi}{3}+ 2n\pi)$. This converges to $0$ $\endgroup$ – miracle173 Nov 20 '15 at 7:02

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